Answer: To find the percentage of people with an IQ score less than 117, we need to calculate the z-score first. The z-score measures how many standard deviations an individual score is from the mean in a normal distribution.
The z-score formula is given by:
z = (x - μ) / σ
Where:
- x = IQ score (117 in this case)
- μ = mean IQ score (100)
- σ = standard deviation (15)
Let's calculate the z-score:
- z = (117 - 100) / 15
- z = 17 / 15
- z ≈ 1.1333
Now, we need to find the percentage of people with a z-score less than 1.1333. We can look up this value in the standard normal distribution table (also known as the Z-table) or use statistical software/tools.
Using the Z-table, we find that the percentage of people with a z-score less than 1.1333 is approximately 0.8708, or 87.08% (rounded to the nearest hundredth).
Therefore, approximately 87.1% of people have an IQ score less than 117.